# Monday, September 12

:::{.remark}
Question from last time: does $L$ always factor as $\Rad(L) \oplus L_{\ss}$ with $L_{\ss}$ semisimple?
Not always, instead there is a semidirect product decomposition $L = \Rad(L) \semidirect \lies$ where $\lies$ is the *Levi subalgebra*.
Consider $L = \liegl_n$, then $\Rad(L) \neq \lieh$ since $[h, e_{ij}] = (h_i - h_j)e_{ij}$, so in fact this forces $\Rad(L) = \CC I_n = Z(L)$ with complementary subalgebra $\lies = \liesl_n$.
Note that $\liegl_n = \CC I_n \oplus \liesl_n$ where $\liesl_n = [L L]$ is a direct sum, and $\liegl_n$ is **reductive**.
:::

## 5.3: Inner Derivations

:::{.theorem title="In semisimples, every derivation is inner"}
If $L$ is semisimple then $\ad(L) = \Der L$.
:::

:::{.proof title="?"}
We know $\ad(L) \leq \Der L$ is a subalgebra, and $L$ semisimple implies $0 = Z(L) = \ker \ad$, so $\ad: L\iso \ad(L)$ is an isomorphism and $\ad(L)$ is semisimple.
The Killing form of a semisimple Lie algebra is always nondegenerate, so $\kappa_{\ad(L)}$ is nondegenerate, while $\kappa_{\Der L}$ may be degenerate.
Recall that $\ad(L) \normal \Der L$, so $[\Der L, \ad(L)] \subseteq \ad(L)$.
Define $\ad(L)^\perp \normal \Der L$ to be the orthogonal complement in $\Der(L)$ with respect to $\kappa_{\Der L}$, which is an ideal by the associative property.

Claim: $\ad(L) \intersect \ad(L)^\perp = 0$, 
This follows readily from the fact that $\kappa_{\ad(L)}$ is nondegenerate and so $\Rad(\kappa_{\ad(L)}) = 0$.

Note that $\ad(L), \ad(L)^\perp$ are both ideals, so $[\ad(L), \ad(L)^\perp] \subseteq \ad(L) \intersect \ad(L)^\perp = 0$.
Let $\delta \in \ad(L)^\perp$ and $x\in L$, then $0 = [\delta, \ad_x] = \ad_{ \delta(x) }$ where the last equality follows from an earlier exercise.
Since $\ad$ is injective, $\delta(x) = 0$ and so $\delta = 0$, thus $\ad(L)^\perp = 0$.
So we have $\Rad \kappa_{\Der L} \subseteq \ad(L)^\perp = 0$ since any derivation orthogonal to all derivations is in particular orthogonal to inner derivations, and thus $\kappa_{\Der L}$ is nondegenerate.
Finally, we can write $\Der L = \ad(L) \oplus \ad(L)^\perp = \ad(L) \oplus 0 = \ad(L)$.
:::

## 5.4: Abstract Jordan Decompositions

:::{.remark}
Earlier: if $A\in \Alg\slice\FF^\fd$, not necessarily associative, $\Der A$ contains the semisimple and nilpotent parts of all of its elements.
Applying this to $A = L\in \Lie\Alg$ yields $L \cong \ad(L) = \Der L$ ad $\ad_x = s + n \in \ad(L) + \ad(L)$, so write $s = \ad_{x_s}$ and $n = \ad_{x_n}$, then
$\ad_x = \ad_{x_s} + \ad_{x_n} = \ad_{x_s + x_n}$ so $x = x_s + x_n$ by injectivity of $\ad$, yielding a definition for the semisimple and nilpotent parts of $x$.
If $L \leq \liegl(V)$, it turns out that these coincide with the usual decomposition -- this is proved in section 6.4.
:::

## 6.1: Modules (Chapter 6: Complete Reducibility of Representations)

:::{.definition title="$L\dash$modules and representations"}
Let $L \in \Lie\Alg\slice \CC^\fd$, then a **representation** of $L$ on $V$ is a homomorphism of Lie algebras $\phi: L \to \liegl(V)$.
For $V\in\Vect\slice \CC$ with an *action* of $L$, i.e. an operation
\[
L\times V &\to V \\
(x, v) &\mapsto x.v
,\]
is an **$L\dash$module**  iff for all $a,b\in \CC, x,y\in L, v,w\in V$,

- (M1) $(ax+by).v = a(x.v) + b(y.v)$.
- (M2) $x.(av+bw) = a(x.v) + b(x.w)$.
- (M3) $[xy].v = x.(y.v) - y.(x.v)$.

:::

:::{.remark}
An $L\dash$module $V$ is equivalent to a representation of $L$ on $V$.
If $\phi: L \to \liegl(V)$ is a representation, define $x.v \da \phi(x)v \da \phi(x)(v)$.
Conversely, for $V\in \mods{L}$ define $\phi: L\to \liegl(V)$ by $\phi(x)(v) \da x.v$.
:::

:::{.example title="?"}
$L\in \mods{L}$ using $\ad$, this yields the **adjoint representation**.
:::

:::{.definition title="Morphism of $L\dash$modules"}
A **morphism** of $L\dash$modules is a linear map $\psi: V\to W$ such that $\psi(x.v) = x.\psi(v)$ for all $x\in L, v\in V$.
It is an isomorphism as an $L\dash$module iff it is an isomorphism of the underlying vector spaces.[^turns_out_linear]
In this case we say $V, W$ are *equivalent* representations.

[^turns_out_linear]: 
It turns out that the inverse map of vector spaces $\psi\inv: W\to V$ is again a morphism of $L\dash$modules.

:::

:::{.example title="?"}
Let $L = \CC x$ for $x\neq 0$, then 

- What is a representation of $L$ on $V$?
  This amounts to picking an element of $\Endo(V)$.

- When are 2 $L\dash$modules on $V$ equivalent?
  This happens iff the two linear transformations are conjugate in $\Endo(V)$.

Thus representations of $L$ on $V$ are classified by Jordan canonical forms when $V$ is finite dimensional.
:::

:::{.definition title="Submodules, irreducible/simple modules"}
For $V\in \mods{L}$, a subspace $W \subseteq V$ is a **submodule** iff it is in invariant subspace, so $x.w\in W$ for all $x\in L, w\in W$.
$V$ is **irreducible** or **simple** if $V$ has exactly two invariant subspaces $V$ and $0$.
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:::{.warnings}
Note that this rules out $0$ as being a simple Lie algebra.
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:::{.definition title="Quotient modules"}
For $W\leq V \in \mods{L}$ a submodule, the **quotient module** is $V/W$ has underlying vector space $V/W$ with action $x.(v+W) \da (x.v) + W$.
This is well-defined precisely when $W$ is a submodule.
:::

:::{.warnings}
$I\normal L \iff \ad(I) \leq \ad(L)$, i.e. ideals corresponds to submodules under the adjoint representation.
However, irreducible ideals need not correspond to simple submodules.
:::